System of 3 equations 3 unknowns grade 2

Question

$$1=4a+2b+c$$

$$4=9a+3b+c$$

$$\frac{-b^2+4ac}{4a}=-\frac{b}{2a}+1$$

It's all the afternoon I try to solve this...

Can you point me in the right direction or show me how it's done?

I replaced c with $-4a-2b+1$ in the other two...

First post, sorry for any mistake...

Answer 1

multiplying the third equation by $$4a$$ we obtain: $$4ac-b^2=-2b+4a$$ solving this equation for $c$ we get $$c=\frac{b^2-2b+4a}{4a}$$ plugging this in equation (1) and (2): $$1=4a+2b+\frac{b^2-2b+4a}{4a}$$ $$4=9a+3b+\frac{b^2-2b+4a}{4a}$$ the equation above is equivalent to $$0=a^2+\frac{1}{2}ab+\frac{b^2-2b}{16}$$ from here we get $$a_{1,2}=-\frac{1}{4}b\pm\sqrt{\frac{b}{8}}$$ this can you set in the last equation to get $b$ we obtain the following Solutions $$a=-1,b=8,c=11$$ or $$a=-3,b=18,c=-23$$

Answer 2

Well, we have the following system of equations:

$$ \begin{cases} 4\cdot\text{a}+2\cdot\text{b}+\text{c}=1\\ \\ 9\cdot\text{a}+3\cdot\text{b}+\text{c}=4\\ \\ \frac{4\cdot\text{a}\cdot\text{c}-\text{b}^2}{4\cdot\text{a}}=1-\frac{\text{b}}{2\cdot\text{a}} \end{cases}\tag1 $$

Using the first and the third equation:

$$\frac{\left(1-2\cdot\text{b}-\text{c}\right)\cdot\text{c}-\text{b}^2}{1-2\cdot\text{b}-\text{c}}=1-\frac{2\cdot\text{b}}{1-2\cdot\text{b}-\text{c}}\space\Longleftrightarrow\space\text{c}=1-\text{b}\pm\sqrt{2}\cdot\sqrt{\text{b}}\tag2$$

Where $\sqrt{2}\cdot\sqrt{\text{b}}\ne\text{b}$

Now, using the first and the second equation we can write:

$$1-4\cdot\text{a}-2\cdot\text{b}=4-9\cdot\text{a}-3\cdot\text{b}\space\Longleftrightarrow\space\text{b}=3-5\cdot\text{a}\tag3$$

So, for the first equation we get:

$$4\cdot\text{a}+2\cdot\left(3-5\cdot\text{a}\right)+\left(1-\left(3-5\cdot\text{a}\right)\pm\sqrt{2}\cdot\sqrt{3-5\cdot\text{a}}\right)=1\tag4$$

And from equation $(4)$ you can find that we get the following two solutions for $\text{a}$: $\text{a}=-3,\text{a}=-1$

Answer 3

This make lay out makes me think of polynomials of degree 2.

$p(x) = ax^2 + bx +c\\ p(1) = 1\\ p(2) = 4\\ $

It would be really nice if $p(x) = (x-1)^2$ unfortunately it doesn't fit with the last equation

$9a + 3b + c = 4\\ 4a + 2b + c = 1\\ 5a + b = 3\\ b = 3-5a$

substitute

$4a + 2(3-5a) + c = 1\\ c = 6a - 5$

And now we can attack the 3rd equation entirely in terms of a

$\frac {-(3-5a)^2 + 4a(6a-5)}{4a} = \frac {-(3-5a)}{2a} + 1\\ -(3-5a)^2 + 4a(6a-5) = -2(3-5a) + 4a\\ -9 + 30 a - 25a^2 + 24a^2 - 20a = -6 + 14a\\ a^2 + 4a + 3 = 0\\ (a+3)(a+1) = 0$

$a = -1, b = 8, c = -11\\ a = -3, b = 18, c = -23$